One mole of a gas expands with temperature T such thaht its volume, V=`KT^(2)`, where K is a constant. If the temperature of the gas changes by `60^(@)C` then the work done by the gas is

Explanation

To solve the problem, we need to calculate the work done by one mole of gas that expands with a volume given by \( V = K T^2 \) when the temperature changes by \( 60^\circ C \).

### Step-by-Step Solution:

1. **Identify the relationship between volume and temperature**:
The volume of the gas is given by:
\[ V = K T^2 \]
where \( K \) is a constant.

2. **Differentiate the volume with respect to temperature**:
To find the change in volume \( dV \) with respect to a change in temperature \( dT \), we differentiate:
\[ dV = \frac{d}{dT}(K T^2) = 2K T \, dT \]

3. **Use the ideal gas law to find pressure**:
From the ideal gas law \( PV = nRT \) (where \( n = 1 \) mole), we can express pressure \( P \) as:
\[ P = \frac{RT}{V} \]
Substituting for \( V \):
\[ P = \frac{RT}{K T^2} = \frac{R}{K T} \]

4. **Substitute \( P \) and \( dV \) into the work done formula**:
The work done \( dW \) by the gas during expansion is given by:
\[ dW = P \, dV \]
Substituting the expressions for \( P \) and \( dV \):
\[ dW = \left(\frac{R}{K T}\right) \left(2K T \, dT\right) = \frac{2R}{K T} K T \, dT = 2R \, dT \]

5. **Integrate to find total work done**:
We need to integrate \( dW \) from the initial temperature \( T_1 \) to the final temperature \( T_2 \). The temperature changes by \( 60^\circ C \), which we can express in Kelvin as \( 60 \, K \) (since the change in Celsius is the same as in Kelvin):
\[ W = \int_{T_1}^{T_2} 2R \, dT = 2R \int_{T_1}^{T_2} dT = 2R (T_2 - T_1) \]
Assuming \( T_1 = 0 \, K \) and \( T_2 = 60 \, K \):
\[ W = 2R (60 - 0) = 120R \]

6. **Final result**:
Thus, the work done by the gas is:
\[ W = 120R \, \text{Joules} \]